CALCIReviewProblemsExam2sols

05 000119 11 1 15 for what values of r does the

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Unformatted text preview: the sake of variety let’s solve this using differentials. The problem could be very well solved by writing the tangent line approximation as above. √ d. [2 points] Approximate 1.1 using L(x). Solution: If k and x are as above, then √ 1.1 ≈ L(0.1) = 1.05. ! e [2 edge of cube is found to in the approximation error part (d)? 14. .Thepoints] aWhatwasthe error be 30 cm with a possible from in measurement of 0.1 cm. Estimate the maximum possible error in computing (a) the volume of the cube (b) the surface area of the cube. Solution: Solution: The error is the actual value minus the approximate value which is 1.05 ≈ −0.00119. √ 1.1 − 1 ! 15. For what values of r does the function y = erx satisfies the differential equation y + 5y − 6y = 0? Solution: ! 16. If f (x) = 3 + x + ex , find (f −1 ) (4). Solution: ! 17. A bacteria culture contains 200 cells initially and grows at a rate proportional to its sze. After half an hour the population has increased to 360 cells. (a) Find the number of bacteria after t hours. Page 1 (b) Find the bacteria after 4 hours. (c) Find the rate of growth after 4 hours. (d) When will the population reach 10,000. Page 1 1 Solution: Page 1 ! 18. At what point on the curve y = [ln(x + 4)]2 is the tangent horizontal? Solution: ! 19. (a) Find an equation of the tangent to the curve y = ex that is parallel to the line x − 4y = 1. (b) Find an equation of the tangent to the curve y = ex that passes through the origin. Page 1 1 Solution: 1 1 ! 20. Find the equation of the tangent line to the curve y = 3 arccos(x/2) at (1, π ). Solution: The slope of the line is y evaluated at 1. y = − √ 1 2 · 1/2. Hence plugging 1 for x we get 1−(x/2) √ √ f (1) = − 3. So the equation of the tangent line is y = − 3(x − 1) + π . 21. A cup of hot chocolate has temperature 80◦ C in a room kept at 20◦ C. After half an hour the hot chocolate cools to 60◦ C. (a) Write a model describing the temperature of hot chocolate as a function of time. What is the temperature after another half hour? (b) When will the chocolate have cooled to 40◦ C? Solution: 22. How long will it take an investement to double in value if the interest rate is 6% compounded continuously? Find ! the rate of change of the investment when t = 1. Solution: ! Page 2 23. Bismuth-210 has a half-life of 5 days. A sample population after 1 day is decaying at a rate of 2 mg/day. Find the initial mass of the sample. Write a model describing the population of the Bismuth-210 sample. Solution: 1/2 = e5k . Hence k = 0.2 ln(0.5). So B = B (0)e0.2 ln(0.5)t . Since the decay rate at t = 1 is 1 mg/day we know that dB |t=1 = 2. Taking derivative of B with respect to t we get dt dB = B (0)0.2 ln(0.5)e0.2 ln(0.5)t . dt Plugging t = 1 we get 2 = B (0)0.2 ln(0.5)e0.2 ln(0.5) So B (0) = 2 0.2 ln(0.5)e0.2 ln 0.5 24. Find the critical points of the function. (a) f (x) = x3 + 3x2 − 24x (b) h(p) = p−1 p2 +4 (c) f (θ) = 2 cos θ + sin2 θ (d) f (x) = x ln x (e) f (x) = xe2x 1 Solution: ! 25. Find the local max/min and the absolute max/min of f on the given interval. (a) f (x) = x4 − 2x2 + 3, [−2, 3] (b) f (x) = x x2 +4 , [0, 3] (c) f (x) = x − 2 cos x, [−π, π ] (d) f (x) = x − ln x, [1/2, 2] Solution: ! 26. Show that 5 is a critical number of the function g (x) = 2 + (x − 5)3 but g does not have a local extreme value at 5. Solution: Page 1 115 / Exam 2 (March 20, 2012) page 7 27. Consider the piecewise linear function f (x) graphed below. ! [16 points] Consider the piecewise linear function f (x) graphed below: Page 1 ! 10 f (x) (4, 6) 8 x (10, −6) ! For each function g (x), find the value of g (3): ! a. [4 points] g (For = sinfunction3 g (x), find the value of g (x): x) each [f (x)] Solution: (a) g (x) = sin(f (x)3 ) (b) g (x) = f (x2 ) x g (x) = cos(f (x)3 ) · 3f (x)2 · f (x) 2 g (3) = f (2) 3 (c) g (x) = ln(f (x)) +cos(7 ) · 3 · 7 · (−1) = 124.0442. (d) g (x) = f −1 (x) b. [4 points] g (x) = f (x2 ) x Solution: g (x) = x · f (x2 ) · 2x − f (x2 ) x2 ! ! For each function g(x), find the value of g (3): ! ! a. [4 points] g(x) = sin [f (x)]3 Solution: g (x) = cos(f (x)3 ) · 3f (x)2 · f (x) g (3) = cos(73 ) · 3 · 72 · (−1) = 124.0442. b. [4 points] g(x) = f (x2 ) x Solution: x · f (x2 ) · 2x − f (x2 ) x2 3(−3)6 − (−3) = −5.667. g (3) = 9 g (x) = c. [4 points] g(x) = ln(f (x)) + f (2) Solution: g (x) = 1 f (x) + 0 f (x) g (3) = −1 1 · (−1) = . 7 7 d. [4 points] g(x) = f −1 (x) Solution: g (x) = g (3) = ! ! 1 f (f −1 (x)) 1 2 =− . f (6) 3 !...
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